The mountain pass theorem and applications.

Thesis Overview

<p> </p><p>Dierentiability in Banach Spaces<br>We will dene here two types of dierentiability in Banach spaces<br>as generalizations of the concept of dierentiability in R.<br>1.1 Gâteaux Derivative<br>Let us denote by B(X; Y ) the space of all bounded linear maps from<br>X to Y where X; Y are Banach spaces.<br>Recall that a bounded linear map means a continuous linear map.<br>Denition1.1<br>Let f : U 7! Y be a mapping and x 2 U ; where U X open. We<br>say that f is Gâteaux dierentiable at xo if there exists A 2 B(X; Y ),<br>such that<br>8 h 2 X n f0g, the map t 7! f(x0+th)ô€€€f(x0)<br>t has a limit as t ! 0<br>equal to A(h); that is,<br>lim<br>t!0<br>f(x0 + th) ô€€€ f(x0)<br>t<br>= A(h) (1:2)<br>or equivalently<br>f(x0 + th) ô€€€ f(x0) = tA(h) + o(t) 8 h 2 X<br>where o(t) holds for the remainder r(t) = f(x0+th)ô€€€f(x0)ô€€€tA(h)<br>6<br>satisfying<br>lim<br>t!0<br>kr(t)k<br>t<br>= 0 :<br>For simplicity we will write Ah instead of A(h):<br>Ah is called the Gâteaux derivative of f at x0 in the direction of<br>h denoted @f<br>@h (x0).<br>The bounded linear operator A, depending on x0, is denoted by<br>DGf(xo) or f0G<br>(xo) and called the Gâteaux dierential.<br>Remarks.<br>– In Denition 1.1, one can simply require that t ! 0+.<br>– Whenever h 6= 0 and the ratio f(x0+th)ô€€€f(x0)<br>t has a limit in Y as<br>t ! 0, we say that f is dierentiable in the direction of h at xo,<br>and we call lim<br>t!0<br>f(x0 + th) ô€€€ f(x0)<br>t<br>the directional dérivative of<br>f at xo in the direction h.<br>Example 1.1:<br>The function f : R2 7! R dened by<br>f(x; y) = x2 + y2<br>is Gateaux dierentiable at every point (x0; y0) 2 R2.<br>Indeed: Let u0 = (x0; y0) and h = (h1; h2). Then<br>f(u0 + th) ô€€€ f(u0) = 2t(x0h1 + y0h2) + t2(h21<br>+ h22<br>); 8t 2 R:<br>It follows that<br>lim<br>t!0<br>f(u0 + th) ô€€€ f(u0)<br>t<br>= 2(x0h1 + y0h2) = 2hu0; hi<br>et since the map h 7! 2hu0; hi is linear and continuous from R2<br>to R, we conclude that f is Gâteaux dierentiable and<br>DGf(u0)(h) = 2hu0; hi 8 hR2:<br>Moreover by regarding R2 as a euclidean space, we can derive the<br>gradient of f at uo as<br>rf(uo) = 2uo :<br>which is actually linear and bounded with respect to h as the inner<br>product( since RN is an inner product space).<br>7<br>Theorem 1.1: (Euler necessary condition for extrema)<br>Let X and Y be real Banach spaces, f : U 7! Y be a mapping and<br>x 2 U where U X is open. If f is Gâteaux dierentiable at an extremum<br>point x0 (maximum or minimum point), then DGf(x0) = 0<br>Proof : Under the hypothesis of this theorem, suppose without loss<br>of generality that x0 is a minimum point (otherwise consider the<br>function ô€€€f instead of f).<br>Since xo 2 U and U is open, there exists a positive real number<br>r such that the open ball Br(xo) is contained in U. Now let<br>h 2 X n f0g. Then for every t such that jtj &lt; r=jjhjj, we have,<br>f(xo + th) f(xo) and so by the Gâteaux dierentiability of f at<br>xo, we have<br>DGf(xo)(h) = lim<br>t!0+<br>f(u0 + th) ô€€€ f(u0)<br>t<br>0<br>It also follows that DGf(xo)(ô€€€h) 0, i.e. DGf(xo)(h) 0 by linearity<br>of DGf(xo). Therefore DGf(xo)(h) = 0 for all h 2 X indeed.<br>Thus DGf(xo) = 0.<br>Theorem 1.2: (Mean Value Theorem in Banach Spaces)<br>Let X and Y be Banach spaces, U X be open and let f : U ! Y<br>be Gâteaux dierentiable. Then for all x1 x2 2 X, we have<br>kf(x1) ô€€€ f(x2)k sup<br>t2[0;1]<br>kDGf(x1 + t(x2 ô€€€ x1)k kx1 ô€€€ x2k<br>provided that the sup<br>t2[0;1]<br>kDGf(x1 + t(x2 ô€€€ x1)k is nite.<br>Proof. Suppose that the assumptions of Theorem 1.2 hold. Let<br>g 2 Y (the dual of Y ) such that jjgjj 1. Then the real-valued<br>function ‘ : [0; 1] ô€€€! R dened by<br>‘(t) = g f(x1 + th) where h = x2 ô€€€ x1<br>is dierentiable on [0; 1 in the usual sense. Moreover we see that<br>‘0(t) = gô€€€<br>DGf(x1 + th)(h)</p><p>; 8 t 2 (0; 1) :<br>It follows from the classical mean valued theorem that<br>j'(1) ô€€€ ‘(0)j sup<br>0j’0(t)j ;<br>that is<br>kg f(x1) ô€€€ g f(x2)k sup<br>0j’0(t)j :<br>8<br>Moreover for all t 2 (0; 1), we have<br>j’0(t)j =</p><p>g<br>ô€€€<br>DGf(x1 + th)(h)</p><p>jjgjj kDGf(x1 + th)k khk<br>kDGf(x1 + th)k khk:<br>And so<br>kgô€€€<br>f(x1)ô€€€f(x2)</p><p>k = kgof(x1)ô€€€gf(x2)kj</p><p>sup<br>0kDGf(x1 + th)k</p><p>khk :<br>But it is well known as a consequence of the Hahn-Banach theorem<br>that<br>kyk = supfu(y) ; u 2 Y ; kuk 1 g:<br>Therefore we nally have<br>kf(x1) ô€€€ f(x2)k sup<br>t2[0;1]<br>kDGf(x1 + t(x2 ô€€€ x1)k kx1 ô€€€ x2k :<br>Remark. If f satises the assumptions of Theorem 1.2 and has a<br>continuous Gâteaux dierential, then one can prove the conclusion<br>of Theorem 1.2 by using the notion of Riemann integration in Banach<br>spaces following the next lemma.<br>Lemma. (Cf. ) Let X be a Banach space and ‘ : [a; b] ! X<br>be continuous, where ô€€€1 &lt; a &lt; b &lt; +1. Then the sequence of<br>partial sums<br>b ô€€€ a<br>n<br>Xnô€€€1<br>k=0<br>‘</p><p>a + k<br>b ô€€€ a<br>n</p><p>converges as n ! 1;<br>and its limit is called the Riemann integral of f over [a; b] and is<br>denoted by Z b<br>a<br>‘(t) dt :<br>That is<br>Z b<br>a<br>‘(t)dt = lim<br>n!1<br>b ô€€€ a<br>n<br>Xnô€€€1<br>k=0<br>‘</p><p>a + k<br>b ô€€€ a<br>n</p><p>:<br>It is easily seen that</p><p>Z b<br>a<br>‘(t) dt</p><p>Z b<br>a<br>k'(t)k dt:<br>9<br>Furthermore if ‘ is continuously (Gâteaux) diérentiable on [a; b] ,<br>then<br>‘(b) = ‘(a) +<br>Z b<br>a<br>‘0(t) dt :<br>Whenener the convergence in (1.2) is uniform for h, there arises<br>an interesting stronger type of dierentiability called the Fréchet<br>dierentiability.<br>1.2 Fréchet Dierentiability<br>It is a rened notion dierentiabilty of which concept is implicitly<br>closer to that of the standard notion of dierentiability known in R<br>Recall: A function f : R ô€€€! R is said to be dierentiable at<br>x0 2 R if and only if the mapping dened on R n f0g as<br>h 7!<br>f(x0 + h) ô€€€ f(x0)<br>h<br>has a limit a 2 R as h ! 0; that is,<br>lim<br>h!0<br>f(x0 + h) ô€€€ f(x0)<br>h<br>= a 2 R:<br>Obeserve that this condition is equivalent to the existence of a real<br>number a 2 R such that<br>f(x0 + h) = f(x0) + ah + o(h) :<br>Now, how can we extend this notion to operators dened between<br>Banach spaces? The answer is in the following denition.<br>Denition 1.2:<br>A function f : U ô€€€! Y ; where Xand Y are Banach spaces and U<br>open in X, is said to be Fréchet dierentiable at a point x0 2 U, if<br>there exists a bounded linear map A : X ! Y such that:<br>lim<br>khk!0<br>kf(x0 + h) ô€€€ f(x0) ô€€€ Ahk<br>khk<br>= 0 ; (1:1)<br>or equivalently<br>f(x0 + h) ô€€€ f(x0) = Ah + o(khk) ; (1:2)<br>10<br>where<br>r(h) := f(x0 + h) ô€€€ f(x0) ô€€€ Ah = o(h)<br>in the sense that<br>lim<br>khk!0<br>kr(h)k<br>khk<br>= 0 :<br>Such an operator A is unique and called the Fréchet dierential<br>of f at x0 and is denoted by Df(x0) or f0(x0) (somtimes it is also<br>denoted by df(xo)).<br>The function f is said to be Fréchet dierentiable (or simply differentiable)<br>on U, if it is Fréchet dierentiable at every point of U:<br>When there is no ambiguity about the domain of f, we just say that<br>f is dierentiable.<br>Denition 1.3:<br>Let X and Y be Banach spaces, U open in X and let f : U ! Y<br>be Fréchet dierentiable on U. The Fréchet dierential of f on U is<br>the mapping<br>Df : U ! B(X; Y )<br>x 7! Df(x) :<br>We say that f is continuously dierentiable on U or a mapping of<br>class C1 (or simply a C1-mapping) if Df is continuous as a mapping<br>from U into B(X; Y ).<br>Examples (Fréchet dierentiable functions).<br>Let H be a real Hilbert space. Then the function F : H ô€€€! R<br>dened by<br>F(x) =<br>1<br>2<br>kxk2<br>is Fréchet dierentiable on H and its Fréchet dierential is dened<br>by:<br>DF(x)(h) = hx; hi = hh; xi:<br>Thanks to the Riesz representation we can write<br>rF(x) = x :<br>11<br>Indeed let us x xo 2 H arbitrarily. Then for every h 2 H, we have<br>F(xo + h) ô€€€ F(xo) = 1<br>2 (kxo + hk2 ô€€€ kxok2)<br>= 1<br>2 (hxo + h; xo + hi ô€€€ hxo; xoi)<br>= 1<br>2 (hxo; xoi + 2hxo; hi + hh; hi ô€€€ hxo; xoi)<br>= hxo; hi + hh;hi<br>2<br>= hh; xoi + jjhjj2<br>2 :<br>Now dene the operator A : H ! H by A(h) = hh; xoi. Then<br>A is linear (since the real inner product is bilinear) and bounded<br>(according to Cauchy-Schwarz inequality). Moreover it is clear that<br>lim<br>khk!0<br>kF(x0 + h) ô€€€ F(x0) ô€€€ A(h)k<br>khk<br>= lim<br>khk!0<br>khk=2 = 0 :<br>Next we present some properties of the Fréchet dierential.<br>Proposition 1.1: Let X and Y be Banach spaces and U X<br>open.<br>1. If F : U ! Y is Fréchet dierentiable at some point x0 2 U,<br>then F is continuous at x0.<br>2. If F : U ! Y is Fréchet dierentiable according to a norm in<br>X,<br>then it is also Fréchet dierentiable according to any norm<br>equivalent to the rst norm.<br>3. (linéarity)<br>If F;G : U ! Y are Fréchet dierentiable at some point xo 2<br>U,<br>then for any a; b 2 R, aF + bG is Fréchet dierentiable at xo<br>and<br>D(aF + bG)(xo) = aDF(xo) + bDG(xo) :<br>4. (Chain rule).<br>Let also V be an open set of a Banach space Z and consider<br>two mappings F : U ô€€€! Y and G : V ô€€€! Z such that<br>F(U) V . If F is Frechet dierentiable at some point xo 2 U<br>12<br>and G : V ô€€€! Z is Frechet dierentiable at yo = F(xo) 2 V ,<br>then G F is Fréchet dierentiable at xo and<br>D(G F)(xo) = DG(yo) DF(xo) :<br>Proof:<br>1. Suppose that f is Fréchet dierentiable at x0 2 U. Then there<br>exists a bounded linear map A : X ! Y such that:<br>f(x0 + h) ô€€€ f(x0) = Ah + o(khk) :<br>It follows from the continuity of A and the denition of o(h)<br>that<br>lim<br>jjhjj!0<br>jjf(x0 + h) ô€€€ f(x0)jj = 0 ;<br>that is<br>lim<br>h!0<br>ô€€€<br>f(x0 + h) ô€€€ f(x0)</p><p>= 0 in Y<br>or simply lim<br>h!0<br>f(x0 + h) = f(x0).<br>2. Let k:k1 and k:k2 be two equivalent norm in X. Then there<br>exist constant &gt; 0 and &gt; 0 such that<br>kxk1 kxk2 kxk1 ; 8 x 2 X :<br>There a mapping g is dened from an open neighbourhood of<br>0 in (X; k:k1) into Y if and only if is dened from an open<br>neighbourhood of 0 in (X; k:k2) into Y . Moreover for any h 6= 0<br>in the domain of g, we have<br>kg(h)k<br>khk1</p><p>kg(h)k<br>khk2</p><p>kg(h)k<br>khk1</p><p>kg(h)k<br>khk2<br>which implies that<br>lim<br>khk1!0<br>kg(h)k<br>khk1<br>= 0 () lim<br>khk2!0<br>kg(h)k<br>khk2<br>= 0 :<br>3. Let ” &gt; 0. Then by the Fréchet dierentiability of the two<br>functions F and G at xo 2 U, we get (indeed) the existence of<br>&gt; 0 such that for every h 2 X satisfying jjhjj &lt; , we have<br>kF(xo + h) ô€€€ F(xo) ô€€€ DF(xo)(h)k<br>”<br>2(jaj + 1)<br>khk<br>13<br>and<br>kG(xo + h) ô€€€ G(xo) ô€€€ DG(xo)(h)k<br>”<br>2(jbj + 1)<br>khk :<br>Thus we have<br>k(aF+bG)(xo+h)ô€€€(aF+bG)(xo)ô€€€aDF(x)(h)ô€€€bDG(x)(h)k “jjhjj :<br>4. We know that<br>F(xo + h) ô€€€ F(xo) = DF(xo)(h) + o(h) (i)<br>and<br>G(yo + h) ô€€€ G(yo) = DG(yo)(h) + o(h) (ii):<br>Therefore<br>G F)(xo + h) ô€€€ (G F)(xo) = G(F(xo + h)) ô€€€ G(F(xo))<br>= G<br>ô€€€<br>F(xo) + DF(xo)(h) + o(khk)</p><p>ô€€€ G(F(xo))<br>= G(F(xo)) + DG(yo)((DF(xo)(h) + o(h) ô€€€ G(F(xo= DG(yo)DF(xo)(h) + DG(yo)(o(h))<br>= DG(yo)DF(xo)(h) + ~o(h)<br>which gives the result since DG(yo)DF(xo) is a bounded linear<br>map and<br>jj~o(h)jj jjDG(yo)jj jjo(h)jj:<br>Theorem 1.3:<br>Every Fréchet dierentiable function is Gâteaux dierentiable and<br>the dierentials coincide..<br>Proof : Let f : U ô€€€! Y ; where Xand Y are Banach spaces and U<br>open in X, be Fréchet dierentiable at a point x0 2 U. We show<br>that f is Gâteaux dierentiable at xo. Fix any v 2 X n f0g. Then<br>we have f0(xo) 2 B(X; Y ) and<br>lim<br>t!0+</p><p>f(xo+tv)ô€€€f(xo)<br>t ô€€€ f0(xo)(v)</p><p>= lim<br>t!0+<br>jjvjj f(xo+tv)ô€€€f(xo)ô€€€f0(xo)(tv)<br>jjtvjj<br>= lim<br>khk!0<br>jjvjj f(xo+h)ô€€€f(xo)ô€€€f0(xo)(h)<br>jjhjj<br>= 0 :<br>Therefore f is Gâteaux dierentiable at xo and moreover DGf(xo) =<br>f0(xo).<br>14<br>Remark: The converse of Theorem 1.2 is false as it can be seen<br>by the example below.<br>Example 1.2 :<br>The function g : R2 ô€€€! R dened by<br>g(x; y) =<br>8&gt;&lt;<br>&gt;:</p><p>x2y<br>x4+y2<br>4<br>if y 6= 0 ;<br>0 if y = 0<br>is Gâteaux dierentiable at (0; 0) but not Fréchet dierentiable at<br>this point.<br>Indeed, let h = (h1; h2) 2 R2. Then for t &gt; 0, we have<br>g(th) ô€€€ g(0; 0)<br>t<br>=<br>8&lt;<br>:<br>t4h41<br>h22<br>(t2h41<br>+h22<br>)4 if h2 6= 0 ;<br>0 if h2 = 0 ;<br>and so<br>lim<br>t!0+<br>g(th) ô€€€ g(0; 0)<br>t<br>= 0 ;<br>yielding the Gâteaux dierentiability of g at (0; 0) with g0(0; 0) 0.<br>But it is seen that g is not Fréchet dierentiable at (0; 0), according<br>to Theorem 1.3, by considering the perturbations H = (h1; h21<br>) as<br>follows :<br>lim<br>h1ô€€€!0+<br>g(h1; h21<br>) ô€€€ g(0; 0)<br>k(h1; h21<br>)k<br>=<br>1<br>16<br>6= 0 :<br>The next theorem gives a useful sucient condition under which<br>Gâteaux dierentiability implies Fréchet dierrentiability.<br>Theorem 1.4:<br>Let X and Y be Banach spaces, U open and nonempty in X and<br>let f : U ! Y . If f has a continuous Gâteaux dierential, then f<br>is Fréchet dierentiable and f 2 C1(U;R).<br>Proof :<br>Let x 2 U and choose &gt; 0 such that B(x; ) U. Let h 2 X such<br>that jjhjj &lt; . Consider the function ‘ : [0; 1] ô€€€! R dened by :<br>‘(t) = f(x + th) ô€€€ f(x) ô€€€ tDGf(x)(h)<br>Since f is Gâteaux dierentiable, it follows that ‘ is dierentiable<br>and<br>‘0(t) = DGf(x + th)(h) ô€€€ DGf(x)(h) :<br>15<br>By applying the Mean Value Theorem to ‘ we have :<br>j'(1) ô€€€ ‘(0)j sup<br>0j’0(t)j ;<br>that is,<br>kf(x + h) ô€€€ f(x) ô€€€ DGf(x)(h)k sup<br>t2(0;1)<br>kDGf(x + th)(h) ô€€€ DGF(x)(h)k<br>sup<br>t2(0;1)<br>kDGf(x + th) ô€€€ DGF(x)k khk:<br>By continuity of the mapping Df : U ! B(X; Y ), we have<br>lim<br>h!0</p><p>sup<br>t2(0;1)<br>kDGf(x + th) ô€€€ DGf(x)k<br>!<br>= 0 ;<br>and so<br>f(x + h) ô€€€ f(x) ô€€€ DGf(x)(h) = o(h)<br>with Df(x) 2 B(X; Y ).<br>Denition 1.2:<br>Let H be a Hilbert space equipped with inner product h:; :i and<br>f : X ô€€€! R be Fréchet dierentiable. Then the mapping<br>rf : H ô€€€! H<br>x 7! rf(x) ;<br>(where rf(x) is the gradient of f at x) is called a potential operator<br>with a potential f : H ô€€€! R.<br>1.3 Second order derivative<br>Let X and Y be real Banach spaces, U open and nonempty in X,<br>and let f : U ! Y be dierentiable. If<br>f0 : U ô€€€! B(X; Y )<br>is dierentiable, then for every x 2 U,<br>(f0)0(x) 2 L(X; B(X; Y ))<br>and is simply denoted by f00(x) or D2f(x). In this case we say that<br>f is twice dierentiable at x and f00(x) is called the second order<br>16<br>dierential of f at x.<br>Observe that in fact<br>f00(x) : X X ô€€€! Y<br>is bilinear and bounded (i.e., continuous) .<br>We recall that a mapping : X X ô€€€! Y is a bounded bilinear<br>map if:<br>1. 8 (x1; x2) 2 X X, 8 y 2 X and 8 ; 2 R,<br>(x1 + x2; y) = (x1; y) + (x2; y)<br>(y; x1 + x2) = (y; x1) + (y; x2)<br>2. 9K 2 (0; 1) such that<br>k(x1: x2)kY Kkx1kXkx2kX :<br>The norm of such a bounded bilinear map is given by:<br>kk = sup fk(x1; x2)kY ; kx1kX 1 and kx2k 1g<br>Note that, more generally, if E1, E2 and E3 are given three normed<br>linear spaces, we can dene a bounded linear map from E1E2 into<br>E3.<br>The space of bounded bilinear maps from X X into Y is isometric<br>to B(X; B(X; Y )). Indeed the map<br>j : B(X2; Y )) ô€€€! B(X; B(X; Y ))<br>A 7! j(A)<br>where j(A) is such that for all x 2 X and for all y 2 Y ,<br>ô€€€<br>j(A)(x)</p><p>(y) = A(x; y) :<br>Moreover jjj(A)jj = jjAjj.<br>Going back to the setting of the denition of the second order<br>dierential, if f : U ! Y is twice dierentiable, then f0 : U !<br>B(X2; Y )). And if f00 is continuous, we say that f is of class C2<br>and we write f 2 C2(U; Y ).<br>17<br>Furthermore we have the following Taylor formula for x 2 U and h<br>suciently small :<br>f(x + h) = f(x) + f0(x)(h) +<br>1<br>2<br>f00(x)(h; h) + o<br>ô€€€<br>khk2<br>X</p><p>(1:5)<br>that can be established by using a notion of Riemann integration in<br>Y such as<br>f(x + h) = f(x) + f0(x)(h) +<br>Z 1<br>0<br>(1 ô€€€ t)f00(x + th)(h; h) dt :<br>These Taylor expansions give the simplest sucient conditions<br>on a critical a C2 functional to be a local extrema.<br>Proposition 1.4<br>Let X and Y be Banach spaces, U open in X and let f : U ! Y<br>be twice continuously dierentiable. Suppose that xo is a critical<br>point of f.<br>1. If there exists a positive real number such that<br>D2f(xo)(h; h) jjhjj2 ; 8 h 2 X ;<br>Then xo is a local minimum point of f.<br>2. If for every x in a neighbourhood of xo, D2f(x) is positive<br>semidenite (in the sense that<br>D2f(x)(h; h) 0 ; 8 h 2 X );<br>then xo is a local minimum point of f over U.<br>3. If U is convex and for every x 2 U, D2f(x) is positive semidefinite<br>(in the sense that<br>D2f(x)(h; h) 0 ; 8 h 2 X );<br>then xo is a minimum point of f over U. Observe in this case<br>that f is convex on U.<br>For instance if H is a real Hilbert space and b 2 H is given then,<br>the critical point xo of the the functional ‘ dened on H by<br>‘(x) =<br>jjxjj2<br>2<br>+ hb; xi ;<br>is the minimum point (i.e., xo = ô€€€b) .<br>18<br>Note.<br>Let H be a Hilbert space and f : H ! R be twice continuously<br>dierentiable. Then for every x 2 H, there exists (according<br>to the Riesz Representation Theorem) a bounded linear operator<br>A : H ! H which is symmetric and satises<br>D2f(x)(h1; h2) = hAh1; h2i :<br>I<br>19</p> <br><p></p>